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2 years ago

Point M( -3, -2) is rotated 270° counterclockwise and then reflected over the x-axis. What are the coordinates of M'?

Mathematics

1 answer:

White raven [17]2 years ago

4 0

Rotation involves moving a point around a center

The coordinates of M" are (-2,-3)

<h3>How to determine the coordinates of M'</h3>

The pre-image M is given as:

M =(-3,-2)

The rule of 270 degrees counterclockwise rotation is:

$(x,y) \to (y,-x)$

So, we have:

M' = (-2,3)

The rule of reflection over the x-axis is:

$(x,y) \to (x,-y)$

So, we have:

M" = (-2,-3)

Hence, the coordinates of M" are (-2,-3)

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3 years ago

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The differential equation $\frac{dB}{dt}+4B=20$ is a first order separable ordinary differential equation (ODE). We know this because a separable first-order ODE has the form:

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We can rewrite our differential equation in the form of a first-order separable ODE in this way:

$\frac{dB}{dt}+4B=20\\\frac{dB}{dt}=20-4B\\\frac{dB}{dt}=4(5-B)\\\frac{1}{5-B}\frac{dB}{dt}=4$

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$\int\limits {\frac{1}{5-B}} \, dB\\\mathrm{Apply\:u-substitution:}\:u=5-B\\\int\limits {\frac{1}{5-B}} \, dB=\int\limits {\frac{1}{u}} \, dB\\\mathrm{du=-dB}\\-\int\limits {\frac{1}{u}} \,du\\\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)\\-\int\limits {\frac{1}{u}} \,du =-\ln \left|u\right|\\\mathrm{Substitute\:back}\:u=5-B\\-\ln \left|5-B\right|\\\mathrm{Add\:a\:constant\:to\:the\:solution}\\-\ln \left|5-B\right|+C$

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If we want to find the explicit general solution of the differential equation

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$|x|=\left \{ {{x, \:if \>x\geq \>0 } \atop {-x, \:if \>x0}} \right.$

$\left|5-B\right|=e^{-4t+D}\\5-B= \pm \:e^{-4t+D}\\B=5 \pm \:e^{-4t+D}\\B=5\pm \:e^{-4t}\cdot e^{D}\\B=5+Ae^{-4t}$

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And the solution is

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